POJ 2230 Watchcow
Watchcow
This problem will be judged on PKU. Original ID: 2230
64-bit integer IO format: %lld Java class name: Main
If she were a more observant cow, she might be able to just walk each of M (1 <= M <= 50,000) bidirectional trails numbered 1..M between N (2 <= N <= 10,000) fields numbered 1..N on the farm once and be confident that she's seen everything she needs to see. But since she isn't, she wants to make sure she walks down each trail exactly twice. It's also important that her two trips along each trail be in opposite directions, so that she doesn't miss the same thing twice.
A pair of fields might be connected by more than one trail. Find a path that Bessie can follow which will meet her requirements. Such a path is guaranteed to exist.
Input
* Lines 2..M+1: Two integers denoting a pair of fields connected by a path.
Output
Sample Input
4 5
1 2
1 4
2 3
2 4
3 4
Sample Output
1
2
3
4
2
1
4
3
2
4
1
Hint
Bessie starts at 1 (barn), goes to 2, then 3, etc...
Source
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = ;
struct arc {
int to,next;
arc(int x = ,int y = -) {
to = x;
next = y;
}
} e[];
int head[maxn],tot,n,m;
bool vis[];
void add(int u,int v) {
e[tot] = arc(v,head[u]);
head[u] = tot++;
}
void dfs(int u) {
for(int &i = head[u]; ~i; i = e[i].next) {
if(vis[i]) continue;
vis[i] = true;
dfs(e[i].to);
}
printf("%d\n",u);
}
int main() {
int u,v;
while(~scanf("%d%d",&n,&m)) {
memset(head,-,sizeof head);
memset(vis,false,sizeof vis);
for(int i = tot = ; i < m; ++i) {
scanf("%d%d",&u,&v);
add(u,v);
add(v,u);
}
dfs();
}
return ;
}
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